*O llAsiacellStrength of m.->02: A steel specimen 12mm diameter has gauge length 50mm. the steel specimenhad tested via tensile test under maximum load 66KN with elongation 7.5mm,and the yield load of this specimen is 15KN with elongation 2.4mm. Calculate:1- The engineering ultimate stress (ultimate tensile strength), and engineeringstrain at this point.2- The engineering stress and strain at yield point.3- The modulus of elasticity, and the modulus of resilience.4- The final or fracture strain of a steel specimen, if you know that the final lengthof specimen after testing is 58.5mm.5- The true stress and strain for ultimate point.Q3: The yoke-and-rod connection is subjected to a tensile force of 15 kN.Determine the average normal stress in each rod and the average shear stress inthe pin A between the members. Finally, find the shear strain in pin A. Take Gteef 75GPA40 mm15 kN,30 mm30 mn15 kN04: In figure as shown below, assume that a 35mm diameter rivet joint the plateswhich are each 150mm wide.(a) if the allowable stresses 200 MPa for bearing in the plate material and 120MPa for shearing of the rivet, determine the minimum thickness of each plate.(b)Under the conditions specified in part (a) what is the largest average stress inthe plate.P35mm

we are suppose to solve only 3 subparts of a question. please post other parts and other question as…

02: A steel specimen 12mm diameter has gauge length 50mm. the steel specimen had tested via tensile test under maximum load 66KN with elongation 7.5mm, and the yield load of this specimen is 15KN with elongation 2.4mm. Calculate: 1- The engineering ultimate stress (ultimate tensile strength), and engineering strain at this point. 2- The engineering stress and strain at yield point. 3- The modulus of elasticity, and the modulus of resilience. 4- The final or fracture strain of a steel specimen, if you know that the final length of specimen after testing is 58.5mm. 5- The true stress and strain for ultimate point.

02: A steel specimen 12mm diameter has gauge length 50mm. the steel specimen had tested via tensile test under maximum load 66KN with elongation 7.5mm, and the yield load of this specimen is 15KN with elongation 2.4mm. Calculate: 1- The engineering ultimate stress (ultimate tensile strength), and engineering strain at this point. 2- The engineering stress and strain at yield point. 3- The modulus of elasticity, and the modulus of resilience. 4- The final or fracture strain of a steel specimen, if you know that the final length of specimen after testing is 58.5mm. 5- The true stress and strain for ultimate point.

Mechanics of Materials (MindTap Course List)

9th Edition

ISBN:9781337093347

Author:Barry J. Goodno, James M. Gere

Publisher:Barry J. Goodno, James M. Gere

Chapter1: Tension, Compression, And Shear

Section: Chapter Questions

Problem 1.6.7P: A wine of length L = 4 ft and diameter d = 0.125 in. is stretched by tensile forces P = 600 lb. The...

See similar textbooks

Similar questions

A wine of length L = 4 ft and diameter d = 0.125 in. is stretched by tensile forces P = 600 lb. The wire is made of a copper alloy having a stress-strain relationship that may be described mathematically by =18,0001+30000.03(=ksi) in which is nondimensional and has units of kips per square inch (ksi). (a) Construct a stress-strain diagram for the material. (bj Determine the elongation, of the wire due to the Forces P. (c) IF the forces are removed, what is the permanent set of the bar? (d) If the forces are applied again, what is the proportional limit?
A tensile test was conducted on a mild steel& the following data was obtained as follows. Diameter of the steel bar = 3 cm,Gauge length of the bar = 20 cm. Load at elastic limit = 250 kN . Extension at a load of 150kN = 0.21 mm . Maximum load = 380 kN . Total extension = 60 mm . Diameter of the rod at failure = 2. 25 cm Determine: (a) Young’s modulus (b) stress at elastic limit (c)the percentage of elongation (d) Percentage decrease in area.
Example: Convert the change in length data in Table 3-2 to engineering stress and strain and plot a stress-strain curve Homework- help Table 3-2 The results of a tensile test of a 0.505 in. diameter aluminum alloy test bar, initial length (1o) = 2 in. Calculated LTO Load (Ib) Change in Length (in.) Stress (psi) Strain (in./in.) 0.000 1000 0.001 0.0005 4,993 14,978 24,963 34,948 37,445 39,442 39,941 39,691 37,944 3000 0.003 0.0015 5000 0.005 0.0025 7000 0.007 0.0035 7500 0.030 0.0150 7900 0.080 0.0400 8000 (maximum load) 0.120 0.0600 7950 0.160 0.0800 7600 (fracture) 0.205 0.1025
200 stress 150 (MPa) 100 50 0.1 0-2 strain From the tensile stress-strain behavior for the brass specimen the diameter is 12-8 mm Calcalate the folluwing:- %3B 1- The Yield stress. 2-Modulus of elasticity. 3 - The Tensile strength.
The proof stress of a material under tensile loading can be considered as can be considered as equivalent to the ultimate tensile strength for aluminium alloys being only applied to engineering steels during a tensile test the intercept on the stress stain curve for a given strain where a line draw is parallel to the elastic response line. the stress at yield for materials have no defined yield stress
A Brass specimen is subjected to a Tensile test in a laboratory. The following data obtained are: Length of Specimen = 367 mm; Diameter of Specimen = 19 mm; Yield Stress = 124 N/mm?: Ultimate Stress = 217 N/mm?: Fracture Stress = 183 N/mm?: % of Elongation = 50 % % of Reduction in area = 27 %. Determine the elongation at 103 kN, Load at yielding. Maximum load, Load at fracture, Final Length & Final diameter. Take Young's Modulus as 104 GPa.
2. In the stress-strain curve for steel shown below, identify following stresses in ksi and write their values in the space below: 2.1. Yield Point: ksi 2.2. Ultimate Tensile Strength: ksi 60- 400 350 50 300 E 40 250 30 200 150 20 100- 10 50 0. Strain, in/in or m/m J stv A Aa Stress, MPa Stress, ksi
Mild Steel is tested in the laboratory. The data obtained are:Length of Specimen = 320 mmDiameter of Specimen = 30 mmElongation under a load of 20 kN = 0.055 mmLoad at Yield Point = 130 kNMaximum Load = 210 kNLoad at fracture = 115 kNLength of Specimen at failure = 380 mmDiameter of Specimen at failure = 17.5 mm Calculate: (1).Young’s Modulus (2). Yield Stress (3). Ultimate Stress
A ul ? KBIS M 1:11 O photo_2021-0... -> 200 stress 150 (MPa) (00 50 0-2 strain From the tensile stress-strain behavior for the brass specimen d the diamcter is 12-8 mm Calculate the folluding:- 1- The Yield stress. 2- Modulus of elasticity. 3 - The Tensile strength . II
The following stress-strain curve was prepared based on a tensile test of a specimen that had a circular cross-section. The gage diameter of the specimen was 0.25 inches and the gage length was 4 inches. The stress scale of the stress-strain diagram is given with the factor a = 10 ksi. Estimate: (a) The modulus of elasticity. (b) The ultimate strength. (c) The yield strength (0.2% offset). (d) The percent elongation at fracture. 2013 Michael Swanbom STRESS VS. STRAIN BY NC SA 7a bat Sat 2at at 0.05 STRAIN 0.01 0.04 0.06 0.08 0.02 0.03 0.07 0.09 STRESS
yields under a load of 8KN; it attains a max load of 15KN and breaks finally A tensile test is carried out on a bar of mild steel of diameter 20mm. The bar Еxample at a load of 7KN. Estimate yP (i) The tensile stress at the yield point (ii) The ultimate tensile test ii. Max loa (iii) The average stress at the breaking point if the diameter of neck is Omm.
Force P and length change AL data are given in table below for the initial portion of a tension test on 7075-T651 Al alloy. The diameter before testing was 9.07 mm., and the gage length Linitial for t length change measurement was 50.8 mm. What tension force is required to cause yielding in a bar of the same material but with a diameter of 20 mm? P, kN AL mm 7.22 0.0839 14.34 0.1636 21.06 26.8 31.7 34.1 35.0 0.241 0.308 0.380 0.484 0.614 36.0 0.924 36.5 1.279 36.9 37.2 1.622 1.994